Can someone please show me how to solve this problem y^2+72 step by step by factoring a difference of squares so that I have a better understanding? Betore asking, no I do not know what to do.

Question

anonymous · Answer

Are we solving for y?

taepoohwae · Answer

y is squared. Its factoring a difference of a square.

anonymous · Answer

Just bear with me here.
y^2+72=0 (Just pretend)
What do you get when you subtract 72 from both sides?

anonymous · Answer

Okay...I'll answer my question. You get y^2=-72

anonymous · Answer

You then square root both sides. Since you can't have the square root of a negative number, you make -72...(-1)(72)
The square root of negative 1 is also know as i. (i stands for imaginary)
You know have $$\sqrt{y ^{2}}=\sqrt{(-1)(72)}$$ or $$y=i \sqrt{72}$$

anonymous · Answer

Now you find the square root of 72, either using a calculator or factoring 72 out to get 3*3*2*2*2. Now since there is two 3s and two 2s, when you square root it you get (3)(2)(square root of 2) giving you a grand total of y=6 times the $$\sqrt{2}$$

taepoohwae · Answer

The answer is supposed to be as follows (a+b)(a-b)

mertsj · Answer

The problem is that 72 is not a perfect square.

anonymous · Answer

The other way to solve it is to factor it to get (y+square root of 72)(y-square root of 72)

taepoohwae · Answer

Thanks

anonymous · Answer

If it were say, y^2-16, then you would get (y+4)(y-4)